Fibonacci: Are you squared yet?

If you generate the first few values of $G_n$ , you'll quickly recognize that all of them are Fibonacci numbers; in fact, $G_n = F_{2n+1}.$ Thus, with $n = 2017$ , we get $k = 2\cdot 2017 + 1 = \boxed{4035}$ as the answer.

For a more rigorous answer, one must prove the equation $G_n = F_{2n+1}$ . Anyone?

Defining $F_0 = 0$ , the matrix representation of Fibonacci numbers $F_n$ is as follows:

$\begin{aligned} \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \\ \implies \begin{pmatrix} F_{n+1} & {\color{#3D99F6}F_n} \\ {\color{#3D99F6}F_n} & {\color{#3D99F6}F_{n-1}} \end{pmatrix}^2 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2n} = \begin{pmatrix} F_{2n+1} & F_{2n} \\ F_{2n} & {\color{#3D99F6}F_{2n-1}} \end{pmatrix} \end{aligned}$

$\begin{aligned} \implies F_n^2 + F_{n-1}^2 & = F_{2n-1} \\ F_{n+1}^n + F_n^2 & = F_{2n+1} \\ \implies G_n & = F_{2n+1} \end{aligned}$

Therefore, $G_{2017} = F_{4035} \implies k = \boxed{4035}$ .